Frequency Spectrum of Convolution Explained

Frequency Spectrum of Convolution

We are given two signals a(t) and b(t). The output signal is the convolution:

$$y(t) = a(t) * b(t)$$

Using the convolution property of the Fourier Transform:

$$Y(f) = A(f)B(f)$$

1. Spectrum of Signal a(t)

The frequencies present in a(t) are:

$$f = ..., -30,-10,0,10,30,... \text{ kHz}$$

Mathematically:

$$A(f)=\sum_{k} A_k\,\delta(f-10k)$$

2. Spectrum of Signal b(t)

The frequencies present in b(t) are:

$$f = -4, 0, 4 \text{ kHz}$$

Mathematically:

$$B(f)=B_{-1}\delta(f+4)+B_0\delta(f)+B_1\delta(f-4)$$

3. Frequency Spectrum of Convolution

Since

$$Y(f)=A(f)B(f)$$

Substituting:

$$ Y(f)=\left(\sum_k A_k\delta(f-10k)\right) \left[B_{-1}\delta(f+4)+B_0\delta(f)+B_1\delta(f-4)\right] $$

Using the property:

$$\delta(f-a)\delta(f-b)=0 \quad \text{if } a \neq b$$

Only frequencies that appear in both spectra survive.

4. Checking Common Frequencies

A(f) Frequencies	B(f) Frequencies	Match
-30	-4,0,4	No
-10	-4,0,4	No
0	0	Yes
10	-4,0,4	No
30	-4,0,4	No

5. Resulting Spectrum

Only the DC component remains:

$$Y(f) = A_0 B_0 \delta(f)$$

Final Result:
The frequency spectrum of y(t) = a(t) * b(t) contains only:

$$f = 0 \text{ kHz}$$

6. Visualization

A(f):  ... -30  -10   0   10   30 ...
B(f):       -4    0    4
Y(f):            0